Question

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how many moles of hydrogen sulfide are needed to produce 13.5 l of sulfur dioxide according to the following reaction at 0°c and 1 atm?
hydrogen sulfide (g) + oxygen (g) → water (l) + sulfur dioxide (g)
amount = moles
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Explanation:

Step1: Write the balanced chemical equation

$$2H_2S(g)+3O_2(g)\to 2H_2O(l)+2SO_2(g)$$

Step2: Use the ideal gas law \(n = \frac{V}{V_m}\) (at \(0^{\circ}C\) and \(1\ atm\), \(V_m=22.4\ L/mol\)) to find the moles of \(SO_2\)

\(n_{SO_2}=\frac{13.5\ L}{22.4\ L/mol}\approx0.603\ mol\)

Step3: Use the mole ratio from the balanced equation

From the equation, the mole ratio of \(H_2S\) to \(SO_2\) is \(1:1\). So \(n_{H_2S}=n_{SO_2}\)

Answer:

\(0.603\)